Welcome back.
On the last video we came to the conclusion that we could
figure out the volume when we rotate a function about the
x-axis, so let's apply that to an actual exercise.
I'm going to erase everything because I don't want you to
memorize this, because frankly I haven't memorized this.
And if you do, you'll forget it one day and then you
won't know how to do it.
But if you understand why it works then you'll never forget.
As long as you remember basic integration.
Maybe you want to memorize it if your teacher tends to give
you a test that don't have much extra time in them, just
to speed up the process.
But you should know what's going on.
So let me draw the axes again.
That's my y-axis.
That's my x-axis.
And so since our first example was y equals square root of
x, let's stick with that.
And for reasons that might become apparent that tends to
be one of the more typical examples when you rotate
things around axes.
So let me see if I can draw it as well as I drew it last time.
Almost.
OK, so that's y equals square root of x, it's just f of x
this time, I've defined it.
This is the x-axis.
That's the y-axis.
And I'm going to rotate this around the x-axis again.
So I'm going to get a sideways looking cup thing.
And let's say I want to figure out the volume of that cup
between the points 0-- and to make it simple, let's just
say the points 0 and 1.
So essentially we're just going to get a cup, a sideways
cup, it's going to look something like this.
It's going to look something like this.
That's going to be the-- that's a horrible-- the
opening of the cup.
Actually why don't I use the circle tool.
It just dawned on me that I had a circle tool.
So the opening of the cup will look like that.
Actually I could draw it right here.
This would be the opening of the cup.
Well-- you can see sometimes that my videos are
a little unplanned.
There you go.
So that would be the opening of the cup.
This is excellent.
This tool is very well suited for what I'm doing here.
We're rotating around that way.
We're turning that function.
So the cup's going to look like that, so the bottom part of the
cup's going to look like this.
And it's solid, so we want the volume of the whole thing.
In future videos I'm going to show you actually how to figure
out the surface area of the cup, which I find in some
ways more interesting.
So how do we think about that again?
Let's just rederive it, but this time we'll use
a specific equation.
So we just have to figure out what is the volume of one disk
and then sum up all the disks.
So let's say this disk right here-- actually let's just take
this disk at the end point right here that I've already
drawn something for.
So what's the radius of this disk?
The radius of that disk is f of x at that point.
Well f of x at that point is just square root of x.
Radius is equal to square root of x.
And so the area of that disk is going to equal pi r squared.
Well the radius is square root of x, so it equals pi times
square root of x squared.
So it equals pi times x.
That's the area of each disk.
And then if we want the volume, you just have to multiply the
area of that surface times the depth of the disk.
I'm just trying to show.
You can imagine that this is kind of like a quarter and this
is the side of the quarter.
We saw in the last video that depth, that's just a very
small change in x, because we want each disk to be
infinitesimally thin.
So the width is just dx at any point.
So the volume of each disk is equal to the area, which we
just figured out was pi x times the depth, times dx.
That's the volume of each disk.
So the total volume is going to be equal to the
sum of all of these.
That was one disk I drew, then you're going to have another
one here, you're going to have another one here,
another one here.
You're going to have infinitely many, and you want them to be
super, super, super thin so that you get an accurate
measure of the exact volume of this curve.
Otherwise it would just be an approximation, and that's
where we use the integral.
So it will be the integral from.
And my original boundaries were 0 to 1.
The disk we used as an example, this is probably you know the
last disk, so this one will actually have a radius of the
square root of 1, which is 1.
Not that you have to know that, I'm just trying to keep
emphasizing the visualization.
So what will be the integral?
Well we're going to go from 0 to 1, and we're going to sum up
a bunch of these disks, which we've already defined,
so it's pi x dx.
This is looking to be a fairly straightforward integral.
So what's the integral of that?
Pi is just a constant and the antiderivative of x is x to
the 1/2 over-- I'm sorry.
It's x squared over 2.
I've been a little rusty since I last did some
antiderivatives.
So we get x squared, we get pi times x squared over 2.
That's the antiderivative of that.
And then we have to evaluate it at 1 and then subtract
it and evaluate it at 0.
And so what do we have.
We get 1/2 pi, so we get pi over 2 minus 0 pi, minus 0.
So it equals pi over 2.
There we go.
We just figured out the volume of this cup from 0 to 1.
Reasonably interesting.
Let's see if we can do that again to figure out the-- just
to give you another example, just hit the point home-- to
see if we can figure out the volume of a sphere,
the equation for the volume of a sphere.
So what's the equation for a circle?
It's x squared plus y squared is equal to r squared.
And let's write that in terms of y is a function of x, just
so we have something that we can work with the
way we learned it.
So we get y squared is equal to r squared minus x squared, and
then we get y is equal to the square root of r squared
minus x squared.
Actually now that I realize it, I'm going to not do this,
because I think I'm going into too complicated a problem.
I did that on a fly.
But in the next video I will do slightly more complicated
without going to this one, because I probably don't have
time for it I just realized.
Anyway I'll see you in the next video.